Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $p = \dfrac{8x + 8}{x + 9} \div \dfrac{x^2 + 7x + 6}{-6x - 54} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{8x + 8}{x + 9} \times \dfrac{-6x - 54}{x^2 + 7x + 6} $ First factor the quadratic. $p = \dfrac{8x + 8}{x + 9} \times \dfrac{-6x - 54}{(x + 1)(x + 6)} $ Then factor out any other terms. $p = \dfrac{8(x + 1)}{x + 9} \times \dfrac{-6(x + 9)}{(x + 1)(x + 6)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 8(x + 1) \times -6(x + 9) } { (x + 9) \times (x + 1)(x + 6) } $ $p = \dfrac{ -48(x + 1)(x + 9)}{ (x + 9)(x + 1)(x + 6)} $ Notice that $(x + 9)$ and $(x + 1)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -48\cancel{(x + 1)}(x + 9)}{ (x + 9)\cancel{(x + 1)}(x + 6)} $ We are dividing by $x + 1$ , so $x + 1 \neq 0$ Therefore, $x \neq -1$ $p = \dfrac{ -48\cancel{(x + 1)}\cancel{(x + 9)}}{ \cancel{(x + 9)}\cancel{(x + 1)}(x + 6)} $ We are dividing by $x + 9$ , so $x + 9 \neq 0$ Therefore, $x \neq -9$ $p = \dfrac{-48}{x + 6} ; \space x \neq -1 ; \space x \neq -9 $